The given equation is
$$\eqalign{ & \,\,\,\,\,\,\left( {x - b} \right)\left( {x - c} \right) + \left( {x - a} \right)\left( {x - c} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0 \cr & \Rightarrow \,\,3{x^2} - 2\left( {a + b + c} \right)x + \left( {ab + bc + ca} \right) = 0 \cr & \,\,\,\,\,\,{\text{Discriminant}} = 4{\left( {a + b + c} \right)^2} - 12\left( {ab + bc + ca} \right) \cr & \,\,\,\,\,\,\, = 4\left[ {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right] \cr & \,\,\,\,\,\,\, = 2\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \geqslant 0\,\,\forall \,\,a,b,c \cr & \therefore {\text{ Roots of given equation are always real}}{\text{.}} \cr} $$

Let $$\alpha ,{\alpha ^2}$$  be the root of $$3{x^2} + px + 3.$$
$$\eqalign{ & \therefore \,\,\alpha + {\alpha ^2} = - \frac{p}{3}\,\,{\text{and }}{\alpha ^3} = 1 \cr & \Rightarrow \,\,\left( {\alpha - 1} \right)\left( {{\alpha ^2} + \alpha + 1} \right) = 0 \cr & \Rightarrow \,\,\alpha = 1\,\,{\text{or }}{\alpha ^2} + \alpha = - 1 \cr & {\text{If }}\alpha = 1,p = - 6\,\,{\text{which is not possible as }}\,p > 0 \cr & {\text{If }}{\alpha ^2} + \alpha = - 1 \cr & \Rightarrow \,\, - \frac{p}{3} = - 1 \cr & \Rightarrow \,\,p = 3. \cr} $$

$$\eqalign{ & {\text{Given, }}\left| {\left[ x \right] - 2x} \right| = 4 \cr & \Rightarrow \left[ x \right] - 2x = \pm 4 \cr & \Rightarrow - \left[ {\left\{ x \right\} + x} \right] = \pm 4 \cr & {\text{The possible values are }} - 4,\,4,\,3.5,\, - 4.5 \cr & {\text{So, total 4 values exist}}{\text{.}} \cr} $$

$$\eqalign{ & 2 + \left| {{e^x} - 1} \right| = {\left( {{e^x}} \right)^2} - 2{e^x} + 1 = {\left| {{e^x} - 1} \right|^2} \cr & \therefore \,\,{\left| {{e^x} - 1} \right|^2} - \left| {{e^x} - 1} \right| - 2 = 0 \cr & {\text{or, }}\left| {{e^x} - 1} \right| = 2, - 1 \cr & \Rightarrow \,\,\left| {{e^x} - 1} \right| = 2 \cr & \Rightarrow \,\,{e^x} - 1 = 2, - 2 \cr & \Rightarrow \,\,{e^x} = 3, - 1 \cr & \Rightarrow \,\,{e^x} = 3. \cr} $$

For the equation $$p{x^2} + qx + 1 = 0$$    to have real roots
$$\eqalign{ & D \geqslant 0\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,{q^2} \geqslant 4p \cr & {\text{If }}p = 1\,\,{\text{then }}{q^2} \geqslant 4 \cr & \Rightarrow \,\,q = 2,3,4 \cr & {\text{If }}p = 2\,\,{\text{then }}{q^2} \geqslant 8 \cr & \Rightarrow \,\,q = 3,4 \cr & {\text{If}}\,{\text{ }}p = 3\,\,{\text{then }}{q^2} \geqslant 12 \cr & \Rightarrow \,\,q = 4 \cr & {\text{If }}p = 4\,\,{\text{then }}{q^2} \geqslant 16 \cr & \Rightarrow \,\,q = 4 \cr & \therefore \,\,{\text{No}}{\text{. of req}}{\text{. equations}} = 7. \cr} $$

If $$x = n \in Z,$$   the equation is $$2n - 2\left[ n \right] = 1\,\,{\text{or 2}}n - 2n = 1\,\left( {{\text{impossible}}} \right).$$
If $$x = n + k,n \in Z,0 < k < 1$$      then the equation is $$2\left( {n + k} \right) - 2\left[ {n + k} \right] = 1$$
or $$2n + 2k - 2n = 1\,\,{\text{or, }}k = \frac{1}{2}\,\,{\text{and }}n \in Z.$$
Alternate Solution
$$\eqalign{ & 2x - 2\left[ x \right] = 1 \cr & \Rightarrow x - \left[ x \right] = \frac{1}{2} \cr & \Rightarrow \left\{ x \right\} = \frac{1}{2},\,{\text{where}}\left\{ {} \right\}{\text{is fractional part function}} \cr & \therefore \,x = n + \frac{1}{2},\,n \in Z \cr} $$

Quadratic equation with real co-efficients and purely imaginary roots can be considered as
$$\eqalign{ & p\left( x \right) = {x^2} + a = 0\,\,{\text{where }}a > 0\,\,{\text{and }}a \in R \cr & {\text{The }}p\left[ {p\left( x \right)} \right] = 0 \cr & \Rightarrow \,\,{\left( {{x^2} + a} \right)^2} + a = 0 \cr & \Rightarrow \,\,{x^4} + 2a{x^2} + \left( {{a^2} + a} \right) = 0 \cr & \Rightarrow \,\,{x^2} = \frac{{ - 2a \pm \sqrt {4{a^2} - 4{a^2} - 4a} }}{2} \cr & \Rightarrow \,\,{x^2} = - a \pm \sqrt a \,\,i \cr & \Rightarrow \,\,x = \sqrt { - a \pm \sqrt a \,\,i} \, \cr & = \,\,\alpha \pm i\,\beta \,\,\,\,{\text{where }}\alpha {\text{,}}\beta \ne {\text{0}} \cr & \therefore \,\,p\left[ {p\left( x \right)} \right] = 0\,\, \cr} $$
has complex roots which are neither purely real nor purely imaginary.

$$\eqalign{ & y - 2\, = \frac{1}{{4 + \left[ {\frac{1}{{4 + \frac{1}{{4 + .....\infty }}}}} \right]}} = \frac{1}{{4 + \left( {y - 2} \right)}} = \frac{1}{{y + 2}} \cr & \Rightarrow \,\left( {y - 2} \right)\,\left( {y + 2} \right) = 1 \cr & \Rightarrow \,{y^2} - 4 = 1 \cr & \Rightarrow \,{y^2} = 5 \cr & \therefore \,y = \pm \sqrt 5 \,{\text{since}}\,y > 0 \cr & \therefore \,y = \sqrt 5 \cr} $$

$$\lambda $$ and $$\mu $$ are the roots of $${x^2} = 5x - 3{\text{ or }}{x^2} - 5x + 3 = 0$$
$$\eqalign{ & \therefore \lambda + \mu = 5{\text{ and }}\lambda \mu = 3 \cr & \frac{\lambda }{\mu } + \frac{\mu }{\lambda } = \frac{{{{\left( {\lambda + \mu } \right)}^2} - 2\lambda \mu }}{{\lambda \mu }} = \frac{{19}}{3} \cr & \frac{\lambda }{\mu } \cdot \frac{\mu }{\lambda } = 1 \cr} $$
∴ Desired equation is $${x^2} - \frac{{19}}{3}x + 1 = 0$$
$${\text{or, }}3{x^2} - 19x + 3 = 0$$

$$y = - {x^2} + 3x - 2 = \frac{9}{4} - \left( {{x^2} - 3x + \frac{9}{4}} \right) - 2 = \frac{1}{4} - {\left( {x - \frac{3}{2}} \right)^2} \leqslant \frac{1}{4}.$$